Physics
Chapter 10b Fixed-Axis Rotation
Defining Torque
- Torque (\(\tau\)): The rotational counterpart to force. Causes a change in rotational motion.
- Intuitive understanding from opening a door:
- Distance from hinges (pivot): Greater distance means more effective rotation.
- Angle of force: Perpendicular force is most effective.
- Magnitude of force: Larger force means more rapid rotation.
- Direction: For rotation in a plane, torque is either clockwise (negative by convention) or counterclockwise (positive by convention).
Figure 10.31: Door rotation demonstrating torque.
- For a force \(\vec{F}\) applied at point \(P\) with position vector \(\vec{r}\) relative to pivot \(O\): \[\vec{\tau} = \vec{r} \times \vec{F}\]
- Magnitude: \[|\vec{\tau}| = |\vec{r} \times \vec{F}| = rF\sin\theta\] where \(\theta\) is the angle between \(\vec{r}\) and \(\vec{F}\).
- Units: Newton-meters (N \(\cdot\) m).
Torque Components and Lever Arm
- Lever Arm (\(r_{\perp}\)): The perpendicular distance from the pivot \(O\) to the line of action of the force \(\vec{F}\). \[r_{\perp} = r\sin\theta\]
- Magnitude of torque in terms of lever arm: \[|\vec{\tau}| = r_{\perp}F\] A greater lever arm results in a greater magnitude of torque.
- Sign of Torque: Determined by the direction of \(\vec{r} \times \vec{F}\) using the right-hand rule.
- Out of the page (positive \(z\)-axis) \(\implies\) positive torque (counterclockwise).
- Into the page (negative \(z\)-axis) \(\implies\) negative torque (clockwise).
Note
Figure 10.33: Torque on a disk. Torque is maximum when \(\theta = 90^\circ\) and zero when \(\theta = 0^\circ\) or \(180^\circ\).
Example: Calculating Torque on a Rigid Body
Flywheel with three forces acting on it. Radius \(r = 0.5 \text{ m}\).
\(|\vec{F_1}| = 20 \text{ N}\) at \(150^\circ\) from \(\vec{r}\).
\(|\vec{F_2}| = 30 \text{ N}\) at \(90^\circ\) from \(\vec{r}\) (clockwise rotation).
\(|\vec{F_3}| = 30 \text{ N}\) at \(0^\circ\) from \(\vec{r}\) (radial).
Figure 10.35: Three forces acting on a flywheel.
Torque from \(\vec{F_1}\):
\(\theta = 150^\circ\) (from \(\vec{r}\) in counterclockwise direction to \(\vec{F_1}\)).
\(\tau_1 = rF_1\sin\theta = (0.5 \text{ m})(20 \text{ N})\sin(150^\circ) = (0.5)(20)(0.5) = 5.0 \text{ N} \cdot \text{m}\).
(Counterclockwise, so positive).
Torque from \(\vec{F_2}\):
\(\theta = 90^\circ\) (angle from \(\vec{r}\) to \(\vec{F_2}\) for the specific direction). Force causes clockwise rotation.
\(\tau_2 = -rF_2\sin(90^\circ) = -(0.5 \text{ m})(30 \text{ N})(1) = -15.0 \text{ N} \cdot \text{m}\).
Torque from \(\vec{F_3}\):
\(\theta = 0^\circ\) (force is radial).
\(\tau_3 = rF_3\sin(0^\circ) = (0.5 \text{ m})(30 \text{ N})(0) = 0 \text{ N} \cdot \text{m}\).
Net Torque:
\(\tau_{net} = \tau_1 + \tau_2 + \tau_3 = 5.0 \text{ N} \cdot \text{m} - 15.0 \text{ N} \cdot \text{m} + 0 = -10.0 \text{ N} \cdot \text{m}\).
Note
A negative net torque means the flywheel will accelerate clockwise.
\(\vec{F_3}\) produces no torque because it acts along the radius, its line of action passes through the pivot point.
Example: Mass Distribution on a Merry-Go-Round
Father pushes a 50.0-kg merry-go-round (radius 1.50 m, uniform disk) with 250 N force at edge, perpendicular to radius.
Figure 10.38: Father pushes a merry-go-round.
- Angular acceleration (\(\alpha\)) when empty:
- Moment of inertia (solid disk): \(I_{m_g} = \frac{1}{2}MR^2 = \frac{1}{2}(50.0 \text{ kg})(1.50 \text{ m})^2 = 56.25 \text{ kg} \cdot \text{m}^2\).
- Net torque (force perpendicular to radius, no friction): \(\tau = rF = (1.50 \text{ m})(250.0 \text{ N}) = 375.0 \text{ N} \cdot \text{m}\).
- \(\alpha = \frac{\tau}{I_{m_g}} = \frac{375.0 \text{ N} \cdot \text{m}}{56.25 \text{ kg} \cdot \text{m}^2} = 6.67 \text{ rad/s}^2\).
- Angular acceleration (\(\alpha\)) with 18.0-kg child at 1.25 m from center:
- Moment of inertia of child (point mass): \(I_c = m_c r_c^2 = (18.0 \text{ kg})(1.25 \text{ m})^2 = 28.13 \text{ kg} \cdot \text{m}^2\).
- Total moment of inertia: \(I_{total} = I_{m_g} + I_c = 56.25 + 28.13 = 84.38 \text{ kg} \cdot \text{m}^2\).
- Net torque remains the same: \(\tau = 375.0 \text{ N} \cdot \text{m}\).
- \(\alpha = \frac{\tau}{I_{total}} = \frac{375.0 \text{ N} \cdot \text{m}}{84.38 \text{ kg} \cdot \text{m}^2} = 4.44 \text{ rad/s}^2\).
Important
Adding the child increases the moment of inertia, which decreases the angular acceleration for the same applied torque, as expected. This demonstrates how mass distribution affects rotational dynamics.
Example: Rotational Work: A Pulley
String pulled with constant downward force \(F = 50 \text{ N}\) for \(1.0 \text{ m}\) (unwound length).
Pulley: \(R = 0.10 \text{ m}\), \(I = 2.5 \times 10^{-3} \text{ kg} \cdot \text{m}^2\). Pulley starts from rest.
Find angular velocity after string unwinds \(1.0 \text{ m}\).
Figure 10.40: String on a pulley.
- Torque on pulley: Only \(F\) creates torque. \(\tau = FR = (50 \text{ N})(0.10 \text{ m}) = 5.0 \text{ N} \cdot \text{m}\). (Weight of pulley and force from bearings act at the axis, so zero torque.)
- Work Done: The force acts over a linear distance \(d = 1.0 \text{ m}\). \(W = Fd = (50 \text{ N})(1.0 \text{ m}) = 50 \text{ J}\). Alternatively, angular displacement \(\Delta\theta = d/R = 1.0 \text{ m} / 0.10 \text{ m} = 10 \text{ rad}\). \(W = \tau\Delta\theta = (5.0 \text{ N} \cdot \text{m})(10 \text{ rad}) = 50 \text{ J}\).
- Work-Energy Theorem: \(W = \frac{1}{2}I\omega_B^2 - \frac{1}{2}I\omega_A^2\) \(50 \text{ J} = \frac{1}{2}(2.5 \times 10^{-3} \text{ kg} \cdot \text{m}^2)\omega_B^2 - 0\) \(\omega_B^2 = \frac{2 \times 50}{2.5 \times 10^{-3}} = \frac{100}{0.0025} = 40000 \text{ rad}^2/\text{s}^2\) \(\omega_B = \sqrt{40000} = 200.0 \text{ rad/s}\).
